Question

Let \( A = \begin{pmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{pmatrix} \). If \( |A|^2 = 25 \), then \( |\alpha| \) equals to

• A. \( 5^2 \)
• B. 1
• C. \( \frac{1}{5} \)
• D. 5

Hint:

For an upper triangular matrix, the determinant is the product of the diagonal elements. Use the given condition on \( |A|^2 \) to solve for unknowns.

Answer 1

The Correct Option is C

To determine \( |\alpha| \), calculate \( |A| \), the determinant of matrix \( A \). Since \( A \) is upper triangular, its determinant is the product of its diagonal elements:

\(|A| = 5 \times \alpha \times 5 = 25\alpha\)

Given:

\(|A|^2 = 25 \rightarrow (25\alpha)^2 = 25\)

Simplify:

\(625\alpha^2 = 25\)

Divide by 625:

\(\alpha^2 = \frac{25}{625} = \frac{1}{25}\)

Square root:

\(|\alpha| = \frac{1}{5}\)